# Probability and Statistics Questions and Answers – Set Theory of Probability – 2

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This set of Probability and Statistics Assessment Questions and Answers focuses on “Set Theory of Probability – 2”.

1. If P(BA) = p(b), then P(A ∩ B) = ____________
a) p(b)
b) p(a)
c) p(b).p(a)
d) p(a) + p(b)

Explanation: P(B/A) = p(b) implies A and B are independent events
Therefore, P(A ∩ B) = p(a).p(b).

2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) 12
b) 13
c) 16
d) 34

Explanation: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
Therefore, probability = 34.

3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then P(A/B) = ?
a) 1 – P(A /B)
b) $$P\frac{(\bar{A})}{(\bar{B})}$$
c) Not Defined
d) $$\frac{1-P(A \cup B)}{P(\bar{B}}$$

Explanation: From definition of conditional probability we have
$$P(\bar{A}/\bar{B})=\frac{\bar{A}\cap\bar{B}}{P(\bar{B})}$$
Using De Morgan’s Law
=$$\frac{P(\bar{A \cup B})}{(\bar{B}}$$
=$$\frac{1-P(A \cup B)}{P(\bar{B}}$$

4. If A and B are two events, then the probability of exactly one of them occurs is given by ____________
a) P(A ∩ B) + P(A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)

Explanation: The set corresponding to the required outcome is
(A ∩ B) ∪ (A ∩ B)
Hence the required probability is
P(P(A ∩ B) ∪ (A ∩ B)) = (A ∩ B) + P(A ∩ B).

5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(M) + P(N) is?
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate

Explanation: Given : P(M ∪ N) = 0.6, P(M ∩ N) = 0.2
P(M ∪ N) + P(M ∩ N) = P(M) + P(N)
2 – (P(M ∪ N) + P(M ∩ N)) = 2 – (P(M) + P(N))
= (1 – P(M)) + (1 – P(N))
2 – (0.6 + 0.2) = P(M) + P(N)
P(M) + P(N) = 2 – 0.8
= 1.2

6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red?
a) $$\frac{xy-y}{x^2+y^2+2xy-(x+y)}$$
b) $$\frac{xy}{x^2+y^2+2xy-(x+y)}$$
c) $$\frac{y^2-y}{x^2+y^2+2xy-(x+y)}$$
a) $$\frac{xy-y}{x^2+y^2+2xy-(x-y)}$$

Explanation: Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Red=$$\frac{x}{x+y-1}$$
Required porbability=$$\frac{y}{(x+y)}\frac{x}{(x+y-1)}=\frac{xy}{x^2+y^2+2xy-(x+y)}$$

7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is?
a) 0.29
b) 0.37
c) 0.61
d) 0.75

Explanation: Given: p(b) = 0.33, P(c) = 0.42
P(B ∩ C) = 0.12
P(BC) = ?
P(BC) = 1 – P(B ∪ C)
= 1 – p(b) – P(c) + P(B ∩ C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.

8. A coin is biased so that its chances of landing Head is 23. If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is?
a) 427
b) 827
c) 49
d) 29

Explanation: Required probability = 23 x 23 x 13 = 427.

9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 27
b) 57
c) 421
d) 1721

Explanation:Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (HW)
P (HW) = P (H) x P (W)
= (1 – P (H)) x (1 – P (W))
= (1 – 23) x (1 – 37)
= 421.

10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then P (A|B) = ?
a) 0.5
b) 0
c) 0.25
d) 1

Explanation: We know that,
P (AB) = P(AB)/P(B)
= P((A ∪ B)/P(B))
= (1 – P(A ∪ B)) /P(B)
= (1 – 0.5)/0.5
= 1.

11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 34
b) 14
c) 12
d) 16

Explanation:S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 28 = 14.

12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) $$\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}$$
b) $$\frac{C_2^5 * C_2^3}{C_2^8}$$
c) $$\frac{C_1^5 * C_1^3}{C_2^8}$$
d) 0.5

Explanation:Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as red = 5C2 /8C2
Probability of picking both balls as yellow = 3C2 /8C2
∴ required probability $$\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}$$.

Sanfoundry Global Education & Learning Series – Probability and Statistics.